[BZOJ]1602: [Usaco2008 Oct]牧場行走

題目大意:給你一棵樹以及相鄰節點之間的距離，之後詢問2點之間的距離。
用dis數組存根節點到其他節點的距離，則有distance(a,b)=dis[a]+dis[b]-dis[lca(a,b)]*2,所以這題可以用樹上倍增求LCA。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

inline char tc(void)
{
    static char fl[10000+1],*A=fl,*B=fl;
    return A==B?B=(A=fl)+fread(fl,1,10000,stdin),A==B?EOF:*A++:*A++;
}

inline void read(int &a)
{
    a=0;static char c;
    while((c=tc())<'0'||c>'9');
    while(c>='0'&&c<='9')
        a=a*10+c-'0',c=tc();
    return ;
}

struct Edge
{
    int to,cost;
    Edge(int a,int b):to(a),cost(b){}
};
int n,q,deep[1001],x,y,z,dis[1001],fa[1001][12];
vector<Edge>s[1001];

void dfs(int x)
{
    int now=1;
    while(deep[x]>(1<<now))
        fa[x][now]=fa[fa[x][now-1]][now-1],++now;
    for (int i=0;i<s[x].size();++i)
        if(!deep[s[x][i].to])
            fa[s[x][i].to][0]=x,
            dis[s[x][i].to]=dis[x]+s[x][i].cost,
            deep[s[x][i].to]=deep[x]+1,
            dfs(s[x][i].to);
    return ;
}

int lca(int x,int y)
{
    if(deep[x]>deep[y])
        swap(x,y);
    int k=deep[y]-deep[x];
    for (int i=0;i<12;++i)
        if(k&(1<<i))
            y=fa[y][i];
    if(x==y)
        return x;
    for (int i=11;i>=0;--i)
        if(fa[x][i]!=fa[y][i])
            x=fa[x][i],y=fa[y][i];
    return fa[x][0];
}

int main(void)
{
    register int i;
    read(n),read(q);
    for(i=1;i<n;++i)
        read(x),read(y),read(z),s[x].push_back(Edge(y,z)),s[y].push_back(Edge(x,z));
    deep[1]=1;
    dfs(1);
    for (i=1;i<=q;++i)
        read(x),read(y),printf("%d\n",dis[x]+dis[y]-dis[lca(x,y)]*2);
    return 0;
}